Question

If the sum of first n,2n and 3n terms of an AP are S1,S2,S3 respectively, then prove that S3 = 3(S2-S1)​

Answer 1

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

$\\ \\\dashrightarrow \sf S_1=S_n$

$\\\dashrightarrow \ S_2=S_{2n}$

$\\ \dashrightarrow S_3=S_{3n}$

$\;\;\underline{\textbf{\textsf{ To Prove :-}}}$

$\\ \\\dashrightarrow \sf S_3=3(S_2-S_1)$

$\;\;\underline{\textbf{\textsf{Formula Used :-}}}$

$\boxed{\sf{\green{ S_n=\dfrac{n}{2}[2a+(n-1)d]}}}$

$\;\;\underline{\textbf{\textsf{ Proof :-}}}$

Here, we know

$\boxed{\sf{\green{ S_n=\dfrac{n}{2}[2a+(n-1)d]}}}$

Therefore,

$\dashrightarrow\sf S_3=S_{3n} = \dfrac{3n}{2}[2a+(3n-1)d]\\ \\ \\ \dashrightarrow\sf S_2=S_{2n} = \dfrac{2n}{2}[2a+(2n-1)d]\\ \\ \\ \dashrightarrow\sf S_1=S_n = \dfrac{n}{2}[2a+(n-1)d]$

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To prove, we have to find the value of -

${\rm\ S_2-S_1}$

Here,

$\dashrightarrow\sf \bigg[ \dfrac{2n}{2}[2a+(2n-1)d] \bigg] - \bigg[ \dfrac{n}{2}[2a+(n-1)d]\bigg]\\ \\ \\\dashrightarrow\sf \bigg[ \dfrac{2n[2a+(2n-1)d]}{2}\bigg] - \bigg[ \dfrac{n[2a+n-1)d]}{2}\bigg]\\ \\ \\\dashrightarrow\sf \bigg[ \dfrac{4na+4n^2 d-2nd}{2}\bigg] -\bigg[\dfrac{2na+n^2d-nd}{2}\bigg] \\ \\ \\ \dashrightarrow\sf \bigg[\dfrac{4na+4n^2d-2nd -2na-n^2d+nd}{2}\bigg] \ \ \ \ \\\ \\ \\ \dashrightarrow\sf \bigg[\dfrac{(4na-2na)+(4n^2d-n^2d)+(-2nd+nd)}{2}\bigg]\\ \\ \\ \dashrightarrow\sf \bigg[\dfrac{2na+3n^2d-nd}{2}\bigg]\\ \\ \\ \dashrightarrow\sf \bigg[\dfrac{ n(2a+3nd-d)}{2}\bigg]\ \ \ \ \ \ \\\ \\ \\ \dashrightarrow\sf \dfrac{n[2a+(3n-1)d]}{2}\\ \\ \\ \dashrightarrow\sf \dfrac{n}{2}[2a+(3n-1)d]$

$\dashrightarrow\sf S_2-S_1= \dfrac{n}{2}[2a+(3n-1)d]\\ \\ \\ \dashrightarrow\sf 3\big(S_2-S_1\big)= 3\bigg(\dfrac{n}{2}[2a+(3n-1)d]\bigg)$........eq(1)

We got

$\dashrightarrow\sf S_3=S_{3n} = \dfrac{3n}{2}[2a+(3n-1)d]$........eq(2)

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After comparing eq(1) and eq(2), we get

$\sf S_3=3(S_2-S_1)$

$\;\;\underline{\textbf{\textsf{ Hence -}}}$

( Proved)

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